We note first that division by zero is not allowed and that for this reason \(9x-4\) may not be equal to zero and that therefore \(x={{4}\over{9}}\) is not a solution.

We now distinguish two cases, namely \(9x-4>0\) and \(9x-4<0\).
In both cases we multiply the inequality on both sides by \(9x-4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(9x-4>0\), i.e. \(x> {{4}\over{9}}\). Then we get \(7<-3(9x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(27x<5\).
Then, dvision by the coefficient of \(x\)gives \(x < {{5}\over{27}}\).
So we have the following system of inequalities: \(x> {{4}\over{9}}\,\wedge\; x < {{5}\over{27}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(9x-4<0\), i.e. \(x< {{4}\over{9}}\). Then we get \(7>-3(9x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(27x>5\).
Then, division by the coefficient of \(x\) gives \(x > {{5}\over{27}}\).
So we have the following system of inequalities: \(x< {{4}\over{9}}\,\wedge\; x > {{5}\over{27}}\)
and this simplifies to \({{5}\over{27}}\lt x\lt {{4}\over{9}}\).

The solution of the original inequality is \({{5}\over{27}}\lt x\lt {{4}\over{9}}\).